PHP select correct radio button based on the result of a query

Question

I'm relatively new to coding in PHP with MySQL. Here's the problem: I want a radio button selected based off on the result I'm getting from my query. For simplicity purposes, I'm going to only select the field that I need to do the task. Overall, the entire site is a repair system for a local computer shop that I work at, and it is my senior project for a class I'm taking at my college. In the database among other fields, there is a adapterType that has the value of "cust" or "csc" for when customer's bring in their computer (mainly a laptop) and if we are using their power adapter or ours.

The query is

    $query =   "SELECT adapterType FROM repair WHERE repairID = '$repairID'";
    $result = mysqli_query($mysqli, $query);
    list($adapter) = mysqli_fetch_row($result);

If I echo out the value of adapter, I get the value that I want.

On another page, I have a list of repairs and next to each is an update button which loads the repair based off of the repairID. On the repairUpate page, I have various text fields that hold certain values that pertain to that field in the repair table. I also have a the radio buttons for the adapter type. I have

    <input type=\"radio\" name=\"adapter\" value=\"csc\"> CSC A/C
    <input type=\"radio\" name=\"adapter\" value=\"cust\"> Cust A/C

How can I have the correct radio button be selected or "checked" based on the value of $adapter that I get from the query?

I know this was a very long post, but I had to explain everything. Thank you for your patience.

<input type=\"radio\" name=\"adapter\" value=\"csc\" <?php echo ($adaptor=="csc") ? "checked":""?>> CSC A/C <input type=\"radio\" name=\"adapter\" value=\"cust\"<?php echo ($adaptor=="cust") ? "checked":""?>> Cust A/C — user2417483
– user2417483, Commented Dec 3, 2017 at 16:37
Your code is vulnerable to SQL injections. Please learn to use prepared statements instead. — tereško
– tereško, Commented Dec 3, 2017 at 16:54
@tereško while updating to prepared statements is best, saying the OP is open to sql injection is currently speculation as they may be using mysqli_real_escape_string(), or other escaping, on $repairID. — Sean
– Sean, Commented Dec 3, 2017 at 17:02

Sean · Accepted Answer · 2017-12-03 16:41:58Z

2

A simple ternary will allow you to set checked="checked"

$checked = ($adapter == 'csc') ? 'checked="checked"' : '';

Where you have multiple buttons, you can created them in a loop, checking the selected -

$array = array("csc"=> "CSC A/C", "cust" => "Cust A/C");
foreach($array as $value => $name) {
   $checked = ($adapter == $value) ? 'checked="checked"' : '';
   echo "<input type=\"radio\" name=\"adapter\" value=\"{$value}\" {$checked}> {$name}";
}

answered Dec 3, 2017 at 16:41

Sean

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1 Comment

Michael Ziminski Over a year ago

Thank you so much! Before I wrote my post, I saw other answers that were similar to the one you posted, but they didn't work. Yours did though, thank you!

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